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Welcome to our next module in an Intuitive Introduction to Probability

decision-making in an uncertain world.

In this module I want to show you a bunch of applications of probabilities.

As I said at the very beginning of this course,

many people tell me "I hate probability."

And I think that attitude comes from that many people don't really see

cool, real-world applications that are really relevant

in everyday decision-making, or relevant

to the well-being of our societies.

And here I really want to convince you that there are cool applications.

We want to start with some playful applications.

Today in the first lecture we're going to look at the Birthday Problem.

And then later on we get into some more serious stuff

from business or the law.

So, let's just dive into the Birthday Problem.

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Two people having birthday on January 18th

or March 22nd or July 1st.

And then the related question: How many people

do you have to have at this party, so that this probability

of at least one pair of birthday people in the room

is larger than a half, larger than 50%?

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These two questions together give us a Birthday Problem.

Before I tell you how to think about this

here's an in-class quiz question for you.

Let's go back to the second question: How many people do you need

to have in a room so that the probability of at least one pair

is larger than 50%?

Here I have a bunch of options of possible answers.

Exactly one of them is correct.

Please have a go at it, and then afterwards I will tell you

how we can calculate these probabilities.

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Welcome back from the in-class quiz question.

Now I want to show you how we can calculate these probabilities.

I've prepared an Excel sheet called Birthday Problems.

Now, before I show you the calculations in that sheet,

I want to tell you a little bit about the strategy

on how we calculate the probabilities that I mentioned before.

It's actually a really tricky question to think about

how many people may have birthday on the same day

because maybe there's one pair, maybe there are two pairs

maybe there are 3 or more pairs, maybe there are triplets.

There are so many possibilities that it quickly

gets totally overwhelming.

And I can tell you, I have no idea on how to calculate

those probabilities that way.

But I can show you much more elegant, much more simpler way.

Let's use the Complement Rule, remember this was one

of the key rules that we learned about in the every first module.

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What's the opposite of having at least one pair,

or 2 pairs or some more?

Ha! The opposite is very easy, very well defined.

All of us at this party, all the people in this room

having birthday on different days.

And it turns out, that probability can be calculated quite easily.

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So now, come with me to the Excel sheet and I'll show you

how we can calculate that probability.

Here we are now at the spreadsheet for the Birthday Problem

that I prepared

Before we get into the calculation, I need to state my assumptions

which are listed over here in column E.

First, we assume there's no February 29th so we assume

there are only 365 days in the given year.

All of those days are equally likely and we assume

all the people in the room are independent.

That means your birthday doesn't affect mine

our birthdays don't affect anyone else's and so on.

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It get's interesting with Person 2.

This person now has only 364 out of 365 days left, why?

The first person picked the day, Oh, that's my birthday

and now we want the second person to be different

well, that leaves only 364.

And so now I start calculating these probabilities here.

One, anything goes for the first person times 364 divided by 365.

So this number here, the .99726 is the probability that a pair

has birthday on different days.

Now the third person enters the party.

What's the probability that her birthday is different

than of the first 2 people?

Ah! Two days are taken in the year. That means 363 days are left.

So we multiply with 363 divided by 365 and get this probability

a little less that 99.2%.

You now see as more and more people enter the room

the probability of being different gets a little smaller, little smaller

and independence, we multiply all of them.

And so now what happens is that this product

gets smaller and smaller and smaller.

Now the amazing thing here is that the probability

drops below a half already at 23.

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So, as soon as there are 23 people in the room

and assuming our assumptions apply, then the probability of all of us

having birthday on a different day is already below a half.

That now means of 23 people, the probability of at least one pair

is larger than a half.

that's very counter-intuitive and quite surprising.

So often people tell me this can't be true, how can this be?

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Now, I calculated probabilities for many other numbers

including 45, 45 is under 6%

And now for 45, I also prepared a Monte Carlo simulation.

How does this work now? Let's go here to the second sheet.

Here now I randomly pick 45 numbers and that's very easy in Excel.

There's this cool function RANDBETWEEN and then I do 1 SIM

I call on 365 on this Swiss computer, in the United States it's 1 comma 365

whatever it is in your language, and this gives me

a random number between 1 and 365.

I do this for all 45 people.

And then I compare, look at all pairs.

So here I compare Person 2 to Person 1.

I compare Person 3 to Person 1 and 2.

Here I compare Person 10 to the first 9.

If I see a match between Person 8 and 5 as I indicate here

then I get a 1, otherwise the spreadsheet shows me a zero.

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Now at the top here, I count the number of matches we have in our spreadsheet.

And now I can recalculate the sheet, either by clicking a button here

or in a Windows computer, easily F9 and I create 45 new birthdays

a new party, 45 new people, and I check is there a pair?

Look at this, right now I have 2 pairs

and so I guess there's at least 1 match.

I do it again, 3 pairs. I do it again, another 2 pairs.

Two pairs, 4 pairs, wow! Look at this, 3 pairs, 1 pair.

Now, you see, it's hard to hit a zero.

Do it for a while, and eventually you will also occasionally

hit a zero, I can tell you in less than 6% of all times.

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This is a fundamentally different problem. Why?

I fix the day. I said let's look at birthdays today.

This now rules out you and me having birthdays

on the same day of the year that's different than today.

That's in the previous version of the problem that was a hit

that was a match, that was a pair.

And there are so many days, 365, where you and I

could have a birthday together.

Here now, that birthday has to be today.

Now the probability calculation changes.

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Here now, notice the different definition.

Now we are looking at the probability that all of us

do not have a birthday today.

Now, for each of us, the probability of not having a birthday today

is 364 divided by 365, it doesn't decrease.

You and I can have both birthday yesterday.

That's still not a match here, so now the probability

as you see here in my spreadsheet

is 364 divided by 365 to the power of the number of people in the room.

Independence, your probability not today

364 divided by 365 times my birthday, not being today

364 times 365 times and so on for all of them.

And now you may think, oh I need 183 people for this to be 50/50.

Why do people think this? Oh, that's

365 divided by a half is 182.5, so if a 183 people

this will be about a half. Wrong.

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Because from those 183 people, quite a few

will have birthdays on the same days.

And therefore, it's not that we have covered half a year

for sure with 183 people, actually less.

You can now play with this, and the surprising result is that

you need 253 people at your party to have a probability

of more than a half, that at least one of them has a birthday today.

That's a very different calculation.

So let me sum this up one more time.

In this last instance of the problem, the question was

what's the probability that you have someone having

a birthday today, so today and a person's birthday matches.

That's very different than in the original Birthday Problem

at the beginning here, where it can be any day of the year.

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And so I think this helps a lot of people to get over

this counter-intuitive result that you only need 23 people

to have a match on some day of the year.

Not today, but some day of the year.

This wraps up the Excel spreadsheet discussion.

Play around with it a little bit more if you want.

And now let's summarize everything in these slides and return there.

Here we are back from our Excel sheet.

Let me briefly summarize our calculations.

We made the following assumptions: there are 365 days in a year.

So we are ignoring February 29th.

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We assume that all days are equally likely

and finally, that we are all independent of each other.

Your birthday doesn't affect mine, nor anyone else's in the room.

And then we just saw the probability now of the complement

of the original event, namely all of us having birthday on different days

is one big long multiplication.

The first person can have birthday on any given day.

The second person now needs to be different

as the first person has taken a day, for example January 1st.

So the second person has 364 out of 365 days left.

So I multiply 1 times 364 divided by 365.

Now the second person comes along

in addition to the first, so in total the third person.

Now this person cannot have birthday on the first 2 days

that Person 1 and Person 2 had a birthday on.

So that's 363 out of 365 which we multiply

with the previous product and so on.

And now you go on until however many people are in this room.

To summarize, here are the results.

On the one hand with 10 people, there's more than 88% chance

that all of us that are in the room, if there are only 10 of us

have birthday on a different day.

With 23 people, that already drops below a half.

Put differently, for the original event that we talked about

what's the probability of having at least one pair?

With only 23 people, that probability already exceeds a half.

So the answer to that quiz question was 23,

the smallest of the choices given.

Now, notice once you have 60 or more people in the room,

the probability of all of them having different birthdays

drops below 1%.

So, that means next time you are at at a party

you may want to offer that bet.

Yeah, I think you will be able to surprise some people.

And that already brings us to the takeaways.

In addition to this rather surprising,

perhaps counter-intuitive result of the Birthday Problem

we also have seen an application of the Complement Rule.

The Complement Rule sometimes can be very, very helpful

when the calculation of the probability of an event

is very, very tedious, or we think maybe

the Complement Rule allows me easier calculation.